Q.4 Express 0.9999… in the form pq. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.

Recurring Decimal

Image for recurring decimal 0.3

Solution:

x=0.9999…=0.9¯

In this decimal one digit 9 repeats; so, when we multiply by 10

10x=9.9¯

10x=9+x

9x=9

x=1

Now this is happening because the more digits we include in this decimal 0.9¯ the closes we get to 1. For example 0.99999 is much closer to 1 than 0.9. And so 0.9999 … 9 repeated infinitely will be very very close to 1 which is what we are getting.

To think in a different way, we can make the difference between 1 and 0.99… as small as we wish by taking enough 9’s.

Q.5 What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 117? Perform the division to check your answer.

Solution:

When we divide a prime number 17 into 1 there will be a remainder, it will be less than 17, i.e. it will be 1, 2 … 16. Now we take this remainder add 0 to it and then divide again with 17. Now we get a second remainder. Since we cannot have a remainder from before (that will cause the repeating block to restart), there are only 16 possible remainders before they will start repeating. Therefore the decimal expansion of 117 can contain maximum 16 digits in the repeating block.

Q.6 Look at several examples of rational numbers in the form pq(q≠0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

Solution:

A fraction (in simplest form/lowest terms) terminates in its decimal form, if the prime factors of the denominator are only 2’s and 5’s or a product of prime factors, 2’s and 5’s. Otherwise it repeats. In the examples listed below, factor the denominators into the product of prime factors and understand what happens.

Terminating and Repearting DecimalsUnderstanding the terminating or decimal fractions in terms of factorization of denominator.

Terminate

Repeat

12, where 2 in the denominator is the prime factor of (2×1).

13 Here, 3 in the denominator is not the prime factor of 2’s & 5’s.

45, where 5 in the denominator is the prime factor of (5×1).

115 Where, 15 (5×3) in the denominator is not the prime factor of 2’s & 5’s.

310, where 10 in the denominator is the prime factor of (2×5).

730 Where, 30 (2×5×3) in the denominator is not the prime factor of 2’s & 5’s.

425,where 25 in the denominator is the prime factor of (5×5).

166 Where, 66 (11×2×3) in the denominator is not the prime factor of 2’s & 5’s.

316, where 16 in the denominator is the prime factor of (22×2×2).

721 Where, 21 (7×3) in the denominator is not the prime factor of 2’s & 5’s.

710, where 20 in the denominator is the prime factor of (2×2×5).

429 Where, 29 (29) in the denominator is not the prime factor of 2’s & 5s.

This is because during the process of division zeroes are added towards the end and only 2 and 5 are guaranteed to divide all numbers ending in zero.